Let π(x) be the number of prime numbers less than x. If p−q=2k for fixed k, then:
π−1(n)−π−1(m)=2k
We apply Taylor development:
1π′∘π−1(c)(n−m)=2k
But π(x)∼li(x), li(x)=∫x2dtln(t) the integral logarithm of Gauss.
ln(π−1(c))∼2k(n−m)
c≤π(e2k)
Thus we deduce the twin prime number conjecture for k=1.